SOLUTION (Continued)
and
(0.63324 0.77397 )
( 0.70588 0.52941 0.47059 )
(0.70588 0.52941 0.47059 )
DA Da DA Da
DA
DB DB DB DB
DB
DC DC DC DC
DC
DA
TT
DA T
DB
TT
DB T
DC
TT
DC T
= =
= +
= =
=−+ −
= =
= +−
Tλ
jk
Tλ
ijk
Tλ
ijk
C
C
C
Equilibrium Condition with
W= −Wj
0: 0
DA DB DC
FWΣ= + + − =TTT j
Substituting the expressions obtained for
, , and
DA DB DC
TT T
and factoring i, j, and k:
0.70588 0.70588 0
DB DC
TT−+ =
(1)
0.63324 0.52941 0.52941 0
DA DB DC
T T TW+ + −=
(2)
0.77397 0.47059 0.47059 0
DA DB DC
TTT−−=
(3)
using conventional algorithms gives,
119.7 lb
DA
T=
98.4 lb
DB
T=
98.4 lb
DC
T=
PROBLEM 2.87
A 36–lb triangular plate is supported by three wires as shown. Determine
the tension in each wire, knowing that a = 6 in.
SOLUTION
By Symmetry
DB DC
TT=
Free-Body Diagram of Point D:
The forces applied at D are:
, , , and
DB DC DA
TTT P
where
(36 lb) .P= =Pj j
To express the other forces in terms of the unit vectors i, j, k, we write
(16 in.) (24 in.) 28.844 in.
(8in.) (24 in.) (6 in.) 26.0 in.
(8 in.) (24 in.) (6 in.) 26.0 in.
DA DA
DB DB
DC DC
=−=
=−− + =
=−− − =
ij
i jk
i jk
C
C
C
and
(0.55471 0.83206 )
( 0.30769 0.92308 0.23077 )
( 0.30769 0.92308 0.23077 )
DA DA DA DA DA
DB DB DB DB DB
DC DC DC DC DC
DA
TT T
DA
DB
TT T
DB
DC
TT T
DC
= = = −
===−− +
===−− −
Tλ ij
Tλ ijk
Tλ ijk
C
C
C
SOLUTION (Continued)
Equilibrium condition:
0: (36 lb) 0
DA DB DC
FΣ= + + + =TTT j
Substituting the expressions obtained for
, , and
DA DB DC
TT T
and factoring i, j, and k:
(0.55471 0.30769 0.30769 ) ( 0.83206 0.92308 0.92308 36 lb)
(0.23077 0.23077 ) 0
DA DB DC DA DB DC
DB DC
TTT TTT
TT
− − +− − − +
+− =
ij
k
Equating to zero the coefficients of i, j, k:
0.55471 0.30769 0.30769 0
DA DB DC
TTT−− =
(1)
0.83206 0.92308 0.92308 36 lb 0
DA DB DC
TTT− − − +=
(2)
0.23077 0.23077 0
DB DC
TT−=
(3)
Equation (3) confirms that
DB DC
TT
=
. Solving simultaneously gives,
14.42 lb; 13.00 lb
DA DB DC
T TT= = =
PROBLEM 2.88
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AC is 60 N, determine the
weight of the plate.
SOLUTION
We note that the weight of the plate is equal in magnitude to the force P exerted by the support on
Point A.
Free Body A:
0: 0
AB AC AD
FPΣ= + + + =TTT j
We have:
( )
(320 mm) (480 mm) (360 mm) 680 mm
(450 mm) (480 mm) (360 mm) 750 mm
(250 mm) (480 mm) 360 mm 650 mm
AB AB
AC AC
AD AD
=−−+ =
=−+ =
=−− =
ijk
ijk
ijk
C
C
C
Thus:
( )
8 12 9
17 17 17
0.6 0.64 0.48
5 9.6 7.2
13 13 13
AB AB AB AB AB
AC AC AC AC AC
AD AD AD AD AD
AB
TT T
AB
AC
TT T
AC
AD
TT T
AD

= = =−−+


= = =−+

= = =−−


Tλ i jk
Tλ i jk
Tλ i jk
C
C
C
Substituting into the Eq.
0FΣ=
and factoring
, , :i jk
85
0.6
17 13
12 9.6
0.64
17 13
9 7.2
0.48 0
17 13
AB AC AD
AB AC AD
AB AC AD
TTT
T T TP
TTT

−+ +



+− − − +



++− =


i
j
k
Dimensions in mm
SOLUTION (Continued)
Setting the coefficient of i, j, k equal to zero:
:i
85
0.6 0
17 13
AB AC AD
TTT−+ + =
(1)
:j
12 9.6
0.64 0
7 13
AB AC AD
T T TP− − − +=
(2)
:k
9 7.2
0.48 0
17 13
AB AC AD
TTT+ −=
(3)
Making
60 N
AC
T=
in (1) and (3):
85
36 N 0
17 13
AB AD
TT− ++ =
(1′)
9 7.2
28.8 N 0
17 13
AB AD
TT+− =
(3′)
Multiply (1′) by 9, (3′) by 8, and add:
12.6
554.4 N 0 572.0 N
13
AD AD
TT−==
Substitute into (1′) and solve for
:
AB
T
17 5
36 572 544.0 N
8 13
AB AB
TT

= +× =


Substitute for the tensions in Eq. (2) and solve for P:
12 9.6
(544N) 0.64(60 N) (572 N)
17 13
844.8 N
P=++
=
Weight of plate
845 NP= =
PROBLEM 2.89
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AD is 520 N, determine the
weight of the plate.
SOLUTION
See Problem 2.88 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
85
0.6 0
17 13
AB AC AD
TTT−+ + =
(1)
12 9.6
0.64 0
17 13
AB AC AD
T T TP− + − +=
(2)
9 7.2
0.48 0
17 13
AB AC AD
TTT+ −=
(3)
Making
520 N
AD
T=
in Eqs. (1) and (3):
80.6 200 N 0
17
AB AC
TT−+ +=
(1′)
90.48 288 N 0
17
AB AC
TT+ −=
(3′)
Multiply (1′) by 9, (3′) by 8, and add:
9.24 504 N 0 54.5455 N
AC AC
TT−= =
Substitute into (1′) and solve for
:
AB
T
17 (0.6 54.5455 200) 494.545 N
8
AB AB
TT=×+ =
Substitute for the tensions in Eq. (2) and solve for P:
12 9.6
(494.545 N) 0.64(54.5455 N) (520 N)
17 13
768.00 N
P=++
=
Weight of plate
768 NP= =
PROBLEM 2.90
In trying to move across a slippery icy surface, a 175–lb man uses
two ropes AB and AC. Knowing that the force exerted on the man
by the icy surface is perpendicular to that surface, determine the
tension in each rope.
SOLUTION
Free–Body Diagram at A
16 30
34 34
N
= +


N ij
and
(175 lb)W= = −Wj j
( 30 ft) (20 ft) (12 ft)
38 ft
15 10 6
19 19 19
AC AC AC AC AC
AC
AC
TTT
AC
T
−+ −
= = =

= −+−


i jk
Tλ
ijk
C
( 30 ft) (24 ft) (32 ft)
50 ft
15 12 16
25 25 25
AB AB AB AB AB
AB
AB
TTT
AB
T
−+ +
= = =

= −++


i jk
Tλ
ijk
C
Equilibrium condition:
0Σ=F
0
AB AC
+ ++ =T T NW
SOLUTION (Continued)
Substituting the expressions obtained for
, ,,
AB AC
TTN
and W; factoring i, j, and k; and equating each of
the coefficients to zero gives the following equations:
From i:
15 15 16 0
25 19 34
AB AC
TT N− − +=
(1)
From j:
12 10 30 (175 lb) 0
25 19 34
AB AC
TT N+ +− =
(2)
From k:
16 6 0
25 19
AB AC
TT−=
(3)
Solving the resulting set of equations gives:
30.8 lb; 62.5 lb
AB AC
TT
= =
PROBLEM 2.91
Solve Problem 2.90, assuming that a friend is helping the man
at A by pulling on him with a force P = −(45 lb)k.
PROBLEM 2.90 In trying to move across a slippery icy surface,
a 175–lb man uses two ropes AB and AC. Knowing that the force
exerted on the man by the icy surface is perpendicular to that
surface, determine the tension in each rope.
SOLUTION
Refer to Problem 2.90 for the figure and analysis leading to the following set of equations, Equation (3)
being modified to include the additional force
( 45 lb) .= −Pk
15 15 16 0
25 19 34
AB AC
TT N− − +=
(1)
12 10 30 (175 lb) 0
25 19 34
AB AC
TT N+ +− =
(2)
16 6 (45 lb) 0
25 19
AB AC
TT−−=
(3)
Solving the resulting set of equations simultaneously gives:
81.3 lb
AB
T=
22.2 lb
AC
T=
consent of McGraw–Hill Education.
SOLUTION (Continued)
and
(0.63324 0.77397 )
( 0.70588 0.52941 0.47059 )
(0.70588 0.52941 0.47059 )
DA Da DA Da
DA
DB DB DB DB
DB
DC DC DC DC
DC
DA
TT
DA T
DB
TT
DB T
DC
TT
DC T
= =
= +
= =
=−+ −
= =
= +−
Tλ
jk
Tλ
ijk
Tλ
ijk
C
C
C
Equilibrium Condition with
W= −Wj
0: 0
DA DB DC
FWΣ= + + − =TTT j
Substituting the expressions obtained for
, , and
DA DB DC
TT T
and factoring i, j, and k:
0.70588 0.70588 0
DB DC
TT−+ =
(1)
0.63324 0.52941 0.52941 0
DA DB DC
T T TW+ + −=
(2)
0.77397 0.47059 0.47059 0
DA DB DC
TTT−−=
(3)
using conventional algorithms gives,
119.7 lb
DA
T=
98.4 lb
DB
T=
98.4 lb
DC
T=
PROBLEM 2.87
A 36–lb triangular plate is supported by three wires as shown. Determine
the tension in each wire, knowing that a = 6 in.
SOLUTION
By Symmetry
DB DC
TT=
Free-Body Diagram of Point D:
The forces applied at D are:
, , , and
DB DC DA
TTT P
where
(36 lb) .P= =Pj j
To express the other forces in terms of the unit vectors i, j, k, we write
(16 in.) (24 in.) 28.844 in.
(8in.) (24 in.) (6 in.) 26.0 in.
(8 in.) (24 in.) (6 in.) 26.0 in.
DA DA
DB DB
DC DC
=−=
=−− + =
=−− − =
ij
i jk
i jk
C
C
C
and
(0.55471 0.83206 )
( 0.30769 0.92308 0.23077 )
( 0.30769 0.92308 0.23077 )
DA DA DA DA DA
DB DB DB DB DB
DC DC DC DC DC
DA
TT T
DA
DB
TT T
DB
DC
TT T
DC
= = = −
===−− +
===−− −
Tλ ij
Tλ ijk
Tλ ijk
C
C
C
SOLUTION (Continued)
Equilibrium condition:
0: (36 lb) 0
DA DB DC
FΣ= + + + =TTT j
Substituting the expressions obtained for
, , and
DA DB DC
TT T
and factoring i, j, and k:
(0.55471 0.30769 0.30769 ) ( 0.83206 0.92308 0.92308 36 lb)
(0.23077 0.23077 ) 0
DA DB DC DA DB DC
DB DC
TTT TTT
TT
− − +− − − +
+− =
ij
k
Equating to zero the coefficients of i, j, k:
0.55471 0.30769 0.30769 0
DA DB DC
TTT−− =
(1)
0.83206 0.92308 0.92308 36 lb 0
DA DB DC
TTT− − − +=
(2)
0.23077 0.23077 0
DB DC
TT−=
(3)
Equation (3) confirms that
DB DC
TT
=
. Solving simultaneously gives,
14.42 lb; 13.00 lb
DA DB DC
T TT= = =
PROBLEM 2.88
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AC is 60 N, determine the
weight of the plate.
SOLUTION
We note that the weight of the plate is equal in magnitude to the force P exerted by the support on
Point A.
Free Body A:
0: 0
AB AC AD
FPΣ= + + + =TTT j
We have:
( )
(320 mm) (480 mm) (360 mm) 680 mm
(450 mm) (480 mm) (360 mm) 750 mm
(250 mm) (480 mm) 360 mm 650 mm
AB AB
AC AC
AD AD
=−−+ =
=−+ =
=−− =
ijk
ijk
ijk
C
C
C
Thus:
( )
8 12 9
17 17 17
0.6 0.64 0.48
5 9.6 7.2
13 13 13
AB AB AB AB AB
AC AC AC AC AC
AD AD AD AD AD
AB
TT T
AB
AC
TT T
AC
AD
TT T
AD

= = =−−+


= = =−+

= = =−−


Tλ i jk
Tλ i jk
Tλ i jk
C
C
C
Substituting into the Eq.
0FΣ=
and factoring
, , :i jk
85
0.6
17 13
12 9.6
0.64
17 13
9 7.2
0.48 0
17 13
AB AC AD
AB AC AD
AB AC AD
TTT
T T TP
TTT

−+ +



+− − − +



++− =


i
j
k
Dimensions in mm
SOLUTION (Continued)
Setting the coefficient of i, j, k equal to zero:
:i
85
0.6 0
17 13
AB AC AD
TTT−+ + =
(1)
:j
12 9.6
0.64 0
7 13
AB AC AD
T T TP− − − +=
(2)
:k
9 7.2
0.48 0
17 13
AB AC AD
TTT+ −=
(3)
Making
60 N
AC
T=
in (1) and (3):
85
36 N 0
17 13
AB AD
TT− ++ =
(1′)
9 7.2
28.8 N 0
17 13
AB AD
TT+− =
(3′)
Multiply (1′) by 9, (3′) by 8, and add:
12.6
554.4 N 0 572.0 N
13
AD AD
TT−==
Substitute into (1′) and solve for
:
AB
T
17 5
36 572 544.0 N
8 13
AB AB
TT

= +× =


Substitute for the tensions in Eq. (2) and solve for P:
12 9.6
(544N) 0.64(60 N) (572 N)
17 13
844.8 N
P=++
=
Weight of plate
845 NP= =
PROBLEM 2.89
A rectangular plate is supported by three cables as shown.
Knowing that the tension in cable AD is 520 N, determine the
weight of the plate.
SOLUTION
See Problem 2.88 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and
(3) below:
85
0.6 0
17 13
AB AC AD
TTT−+ + =
(1)
12 9.6
0.64 0
17 13
AB AC AD
T T TP− + − +=
(2)
9 7.2
0.48 0
17 13
AB AC AD
TTT+ −=
(3)
Making
520 N
AD
T=
in Eqs. (1) and (3):
80.6 200 N 0
17
AB AC
TT−+ +=
(1′)
90.48 288 N 0
17
AB AC
TT+ −=
(3′)
Multiply (1′) by 9, (3′) by 8, and add:
9.24 504 N 0 54.5455 N
AC AC
TT−= =
Substitute into (1′) and solve for
:
AB
T
17 (0.6 54.5455 200) 494.545 N
8
AB AB
TT=×+ =
Substitute for the tensions in Eq. (2) and solve for P:
12 9.6
(494.545 N) 0.64(54.5455 N) (520 N)
17 13
768.00 N
P=++
=
Weight of plate
768 NP= =
PROBLEM 2.90
In trying to move across a slippery icy surface, a 175–lb man uses
two ropes AB and AC. Knowing that the force exerted on the man
by the icy surface is perpendicular to that surface, determine the
tension in each rope.
SOLUTION
Free–Body Diagram at A
16 30
34 34
N
= +


N ij
and
(175 lb)W= = −Wj j
( 30 ft) (20 ft) (12 ft)
38 ft
15 10 6
19 19 19
AC AC AC AC AC
AC
AC
TTT
AC
T
−+ −
= = =

= −+−


i jk
Tλ
ijk
C
( 30 ft) (24 ft) (32 ft)
50 ft
15 12 16
25 25 25
AB AB AB AB AB
AB
AB
TTT
AB
T
−+ +
= = =

= −++


i jk
Tλ
ijk
C
Equilibrium condition:
0Σ=F
0
AB AC
+ ++ =T T NW
SOLUTION (Continued)
Substituting the expressions obtained for
, ,,
AB AC
TTN
and W; factoring i, j, and k; and equating each of
the coefficients to zero gives the following equations:
From i:
15 15 16 0
25 19 34
AB AC
TT N− − +=
(1)
From j:
12 10 30 (175 lb) 0
25 19 34
AB AC
TT N+ +− =
(2)
From k:
16 6 0
25 19
AB AC
TT−=
(3)
Solving the resulting set of equations gives:
30.8 lb; 62.5 lb
AB AC
TT
= =
PROBLEM 2.91
Solve Problem 2.90, assuming that a friend is helping the man
at A by pulling on him with a force P = −(45 lb)k.
PROBLEM 2.90 In trying to move across a slippery icy surface,
a 175–lb man uses two ropes AB and AC. Knowing that the force
exerted on the man by the icy surface is perpendicular to that
surface, determine the tension in each rope.
SOLUTION
Refer to Problem 2.90 for the figure and analysis leading to the following set of equations, Equation (3)
being modified to include the additional force
( 45 lb) .= −Pk
15 15 16 0
25 19 34
AB AC
TT N− − +=
(1)
12 10 30 (175 lb) 0
25 19 34
AB AC
TT N+ +− =
(2)
16 6 (45 lb) 0
25 19
AB AC
TT−−=
(3)
Solving the resulting set of equations simultaneously gives:
81.3 lb
AB
T=
22.2 lb
AC
T=
consent of McGraw–Hill Education.